Solve for $x$ to the nearest thousandth: $e^{2x}-6e^x+5=0$ .

Step 1: Factor the equation if possible. In this case, the equation is given by: $e^{2x}-6e^x+5=0$. Let's try to see if we can factor this equation. If we can't, we will proceed to use the substitution method to solve the equation. Unfortunately, this equation cannot be factored, so we will use the substitution method. Step 2: Perform the substitution. We will now perform the substitution method by replacing $e^x$ with a variable, say $u$. So, our equation becomes: $u^2-6u+5=0$. This is now a quadratic equation, and we can solve it using the quadratic formula. Step 3: Solve the quadratic equation for u. Now that we have the quadratic equation $u^2-6u+5=0$, we can solve it using the quadratic formula: $u=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-6$, and $c=5$. Plugging in the values, we get: $u=\frac{6\pm \sqrt{(-6{)}^2-4(1)(5)}}{2(1)}$, $u=\frac{6\pm \sqrt{16}}{2}$, So, we have two solutions for $u$: $u_1=\frac{6+\sqrt{16}}{2}=5$, $u_2=\frac{6-\sqrt{16}}{2}=1$. Step 4: Solve for x using the substitution. Now, we need to solve for $x$ using the substitution $u=e^x$. Since we have two solutions for $u$, we will need to solve two equations separately: For $u_1=5$, $5=e^x\Rightarrow x=\ln 5$. For $u_2=1$, $1=e^x\Rightarrow x=\ln 1$. Step 5: Round the answers to the nearest thousandth. Now that we have found the solutions for $x$, we will round the answers to the nearest thousandth: $x_1=\ln 5\approx 1.609$, $x_2=\ln 1=0$. So, the solutions to the equation $e^{2x}-6e^x+5=0$ are $x\approx 1.609$ and $x=0$.